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Sloan [31]
3 years ago
8

Carbon disulfide burns with oxygen to form carbon dioxide and sulfur dioxide in the following manner: CS2 + 3O2 → CO2 + 2SO2

Chemistry
1 answer:
Virty [35]3 years ago
4 0

Answer:

89.6 L.

Explanation:

  • For the balanced reaction:

<em>CS₂ + 3O₂ → CO₂ + 2SO₂.</em>

  • It is clear that <em>1.0 mole of CS₂</em> with 3.0 moles of O₂ to produce 1.0 mole of  CO₂ and <em>2.0 moles of SO₂.</em>

<em></em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of CS₂ produces → 2.0 moles of SO₂.

2.0 mole of CS₂ produces → ??? moles of SO₂.

∴ The no. of moles of SO₂ produced = (2.0 mol)(2.0 mol)/(1.0 mol) = 4.0 mol.

  • It is known that 1.0 mole of any gas at STP occupies 22.4 L.

<em>∴ 4.0 moles of SO₂ occupy = (4.0)(22.4 L) = 89.6 L.</em>

<em></em>

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The island of stability is a term from nuclear physics that describes the possibility of elements with particularly stable "magic numbers" of protons and neutrons. This would allow certain isotopes of some transuranic elements to be far more stable than others, that is, decay much more slowly.

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3 years ago
A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 m
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The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

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HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

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100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

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8 0
2 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
4 years ago
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